If the reaction consumes methane gas ( CH4 ) at a rate of 2.08 M/s, what is the rate of formation of H2 ? the balanced equation is CH4 + N2Cl4 = CCl4 + N2 + 2 H2

4.16M/s

Explanation:

Based on the reaction:

CH₄ + N₂Cl₄ ⇄ CCl₄ + N₂ + 2H₂

1 mole of methane, CH₄, produce 2 moles of H₂.

That means whereas 1 mole of methane is consumed, 2 moles of H₂ are formed

Having this in mind, if you are consuming methane at a rate of 2.08M/s, the rate of formation of hydrogen must be twice this rate, because there are produced twice moles of H₂.

Thus, rate of formation of H₂ is:

2.08M/s ₓ 2 =

4.16M/s